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This question is about metals and the reactivity series - AQA - GCSE Chemistry - Question 2 - 2020 - Paper 1

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This question is about metals and the reactivity series. 02.1 Which two statements are properties of most transition metals? Tick (✓) two boxes. - They are soft ... show full transcript

Worked Solution & Example Answer:This question is about metals and the reactivity series - AQA - GCSE Chemistry - Question 2 - 2020 - Paper 1

Step 1

Which two statements are properties of most transition metals?

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Answer

The two statements that are properties of most transition metals are:

  1. They form ions with different charges.
  2. They have high melting points.

Step 2

Explain how these observations show that silver is less reactive than copper.

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Answer

The pale grey crystals that form are silver, indicating that copper displaces silver from the silver nitrate solution. This happens because copper is more reactive than silver. The solution turning blue shows that copper ions are now in solution, confirming that copper has reacted and displaced less reactive silver.

Step 3

Plan an investigation to identify the three metals by comparing their reactions with dilute hydrochloric acid.

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Answer

  1. Add the metals to dilute hydrochloric acid: Introduce small pieces of magnesium, iron, and copper into separate test tubes containing dilute hydrochloric acid.
  2. Observe reactions: Check for signs of reaction, such as bubbling (indicating hydrogen gas production) or temperature change.
  3. Record results: Note which metals react and which do not. For instance, magnesium will produce a vigorous reaction, iron will show some bubbling, and copper will show no reaction.
  4. Control variables: Ensure that each test uses the same concentration and volume of hydrochloric acid, same mass of each metal, and same particle size for consistency. This plan should yield valid results that help identify each metal based on their reactions with the acid.

Step 4

Calculate the relative atomic mass (A_r) of metal M.

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Answer

To calculate the relative atomic mass (A_r) of metal M, we use the formula: Ar=(203×30)+(205×70)100A_r = \frac{(203 \times 30) + (205 \times 70)}{100} Calculating this gives: Ar=6090+14350100=20440100=204.4A_r = \frac{6090 + 14350}{100} = \frac{20440}{100} = 204.4 Thus, the relative atomic mass of metal M is 204.4.

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